Chapter 11, Problem 9PS

To determine

Prove that triple vector product of given vectors

Answer to Problem 9PS

The vector triple product of vectors u, v and w is $u\times (v\times w)=(u\cdot w)v-(u\cdot v)w$ is verified.

Explanation of Solution

Given:

The three vectors u, v and w are

  $\begin{array}{l}u=\left(u_1,u_2,u_3\right)\hfill \\ v=\left(v_1,v_2,v_3\right)\hfill \\ w=\left(w_1,w_2,w_3\right)\hfill \end{array}$

Consider the vectors in space $a=\langle x_1,y_1,z_1\rangle$ and $\text{b}=\langle x_2,y_2,z_2\rangle$ then the cross product of

  $a=\langle x_1,y_1,z_1\rangle$ and $\text{b}=\langle x_2,y_2,z_2\rangle$ will be the vector

  $\begin{array}{l}a\times b=\left|\begin{array}{ccc}i& j& k\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ =i\left[y_1\left(z_2\right)-y_2\left(z_1\right)\right]-j\left[x_1\left(z_2\right)-x_2\left(z_1\right)\right]+k\left[x_1\left(y_2\right)-x_2\left(y_1\right)\right]\end{array}$

And

  $a\cdot b=\left(x_1\cdot x_2+y_1\cdot y_2+z_1\cdot z_2\right)$

Let three vectors u, v and w are

  $\begin{array}{l}u=\left(u_1,u_2,u_3\right)\hfill \\ v=\left(v_1,v_2,v_3\right)\hfill \\ w=\left(w_1,w_2,w_3\right)\hfill \end{array}$

Vector tripleproductofthesethreevectorsis $u\times (v\times w)$

Cross product of v and w is

  $v\times w=\left(\left(u_2{v}_3-u_3{v}_2\right),\left(u_3{v}_1-u_1{v}_3\right),\left(u_1{v}_2-u_2{v}_1\right)\right)$

Therefore

  $\begin{array}{l}u\cdot (v\times w)=u\cdot \left(\left(v_2{w}_3-v_3{w}_2\right),\left(v_3{w}_1-v_1{w}_3\right),\left(v_1{w}_2-v_2{w}_1\right)\right)\\ =u\cdot \left|\begin{array}{ccc}i& j& k\\ v_1& v_2& v_3\\ w_1& w_2& w_3\end{array}\right|\\ =\left(u_{1}i+u_{2}j+u_{3}k\right)\cdot \left|\begin{array}{ccc}i& j& k\\ v_1& v_2& v_3\\ w_1& w_2& w_3\end{array}\right|\end{array}$

Onperformingfollowingcross product,resultcomes

  $\begin{array}{l}u\times (v\times w)=\left[u_2\left(v_1{w}_2-v_2{w}_1\right)-u_3\left(v_3{w}_1-v_1{w}_3\right)\right]i\\ =+\left[u_3\left(v_2{w}_3-v_3{w}_2\right)-u_1\left(v_1{w}_2-v_2{w}_1\right)\right]j\\ =+\left[u_1\left(v_3{w}_1-v_1{w}_3\right)-u_2\left(v_2{w}_3-v_3{w}_2\right)\right]k\end{array}$

On rearranging above equation

  $\begin{array}{l}u\times (v\times w)=\left[v_1\left(u_2{w}_2+u_3{w}_3\right)i+v_2\left(u_1{w}_1+u_3{w}_3\right)j+u_1\left(u_1{w}_1+u_2{w}_2\right)k\right]\\ =\begin{array}{l}-\left[w_1\left(u_2{v}_2+u_3{v}_3\right)i+w_2\left(u_1{v}_1+u_3{v}_3\right)j+u_2\left(u_1{v}_1+u_2{v}_2\right)k\right]\hfill \\ =(u\cdot w)v-(u\cdot v)w\hfill \end{array}\end{array}$

Hence the expression $u\times (v\times w)=(u\cdot w)v-(u\cdot v)w$ is verified.